RD Chapter 3 Functions Ex 3.2 Solutions
Question - 1 : - If f (x) = x2 – 3x + 4, then find the values of x satisfying the equation f (x) = f (2x + 1).
Answer - 1 : -
Given:
f(x) = x2 – 3x + 4.
Let us find x satisfying f (x) = f (2x + 1).
We have,
f (2x + 1) = (2x + 1)2 –3(2x + 1) + 4
= (2x) 2 +2(2x) (1) + 12 – 6x – 3 +4
= 4x2 + 4x + 1 – 6x+ 1
= 4x2 – 2x + 2
Now, f (x) = f (2x + 1)
x2 – 3x + 4 = 4x2 – 2x + 2
4x2 – 2x + 2 – x2 + 3x – 4 = 0
3x2 + x – 2 = 0
3x2 + 3x – 2x – 2 =0
3x(x + 1) – 2(x + 1) = 0
(x + 1)(3x – 2) = 0
x + 1 = 0 or 3x – 2 = 0
x = –1 or 3x = 2
x = –1 or 2/3
∴The values of x are –1 and 2/3.
Question - 2 : - If f (x) = (x – a)2 (x – b)2, find f (a + b).
Answer - 2 : -
F (x) = (x – a)2(x – b)2
Let us find f (a + b).
We have,
f (a + b) = (a + b – a)2 (a+ b – b)2
f (a + b) = (b)2 (a)2
∴ f(a + b) = a2b2
Question - 3 : - If y = f (x) = (ax – b) / (bx – a), show that x = f (y).
Answer - 3 : -
Given:
y = f (x) = (ax – b) / (bx – a) ⇒ f (y) = (ay – b) / (by – a)
Let us prove that x = f (y).
We have,
y = (ax – b) / (bx – a)
By cross-multiplying,
y(bx – a) = ax – b
bxy – ay = ax – b
bxy – ax = ay – b
x(by – a) = ay – b
x = (ay – b) / (by – a) = f (y)
∴ x= f (y)
Hence proved.
Question - 4 : - If f (x) = 1 / (1 – x), show that f [f {f (x)}] = x.
Answer - 4 : -
Given:
f (x) = 1 / (1 – x)
Let us prove that f [f {f (x)}] = x.
Firstly, let us solve for f {f (x)}.
f {f (x)} = f {1/(1 – x)}
= 1 / 1 – (1/(1 – x))
= 1 / [(1 – x – 1)/(1 – x)]
= 1 / (-x/(1 – x))
= (1 – x) / -x
= (x – 1) / x
∴ f {f (x)} = (x – 1) / x
Now, we shall solve for f [f {f (x)}]
f [f {f (x)}] = f [(x-1)/x]
= 1 / [1 – (x-1)/x]
= 1 / [(x – (x-1))/x]
= 1 / [(x – x + 1)/x]
= 1 / (1/x)
∴ f [f {f (x)}] = x
Hence proved.
Question - 5 : - If f (x) = (x + 1) / (x – 1), show that f [f (x)] = x.
Answer - 5 : -
Given:
f (x) = (x + 1) / (x – 1)
Let us prove that f [f (x)] = x.
f [f (x)] = f [(x+1)/(x-1)]
= [(x+1)/(x-1) + 1] / [(x+1)/(x-1) – 1]
= [[(x+1) + (x-1)]/(x-1)] / [[(x+1) – (x-1)]/(x-1)]
= [(x+1) + (x-1)] / [(x+1) – (x-1)]
= (x+1+x-1)/(x+1-x+1)
= 2x/2
= x
∴ f [f (x)] = x
Hence proved.
Question - 6 : - If
Find:
(i) f (1/2)
(ii) f (-2)
(iii) f (1)
(iv) f (√3)
(v) f (√-3)
Answer - 6 : -
(i) f(1/2)
When, 0 ≤ x ≤ 1, f(x) = x
∴f (1/2) = ½
(ii) f(-2)
When, x < 0, f(x) = x2
f (–2) = (–2)2
= 4
∴ f(–2) = 4
(iii) f(1)
When, x ≥ 1, f (x) = 1/x
f (1) = 1/1
∴ f(1)= 1
(iv) f(√3)
We have √3 = 1.732 > 1
When, x ≥ 1, f (x) = 1/x
∴f (√3) = 1/√3
(v) f(√-3)
We know √-3 is not a real number and the function f(x) isdefined only when x ∈ R.
∴ f(√-3) does not exist.